On the nature of the infinite fall toward the EH

[PeterDonis]

- closed manifolds are rejected; null infinity must be well defined.

The problem with this is that it doesn't just rule out closed manifolds; it rules out all manifolds that aren't asymptotically flat. You can only define null infinity in an asymptotically flat manifold. For example, an open FRW manifold such as the one currently used in the "best fit" model for our universe has no null infinity.

A less artificial way to change GR is to add evaporation to it in such a way as to guarantee that no event loses connection to null infinity before evaporation completes.

This doesn't necessarily have to change GR; you could (I believe) construct the classical limit of such an "evaporation" model by using an SET with a sufficiently large negative pressure. This would violate several energy conditions, so such SETs are usually considered "unphysical", but when quantum effects are included it's no longer clear that the energy conditions always have to hold anyway.

 

[PeterDonis]

I, of course, feel that there is no physical basis for an additional law like this - it only serves to violate the equivalence principle.

Not only that, it appears to require that local physics--whatever it is that, locally, prevents an event horizon from forming--must "know" the entire future of the spacetime, so that local events can "know" when they are getting close to losing connection with null infinity (more precisely, with *future* null infinity, which is the point).

 

[PAllen]

The problem with this is that it doesn't just rule out closed manifolds; it rules out all manifolds that aren't asymptotically flat. You can only define null infinity in an asymptotically flat manifold. For example, an open FRW manifold such as the one currently used in the "best fit" model for our universe has no null infinity.

I'm not sure that's right. The technical definition of horizon uses null infinity. I've seen claims in the literature that BH's are only technically undefinable for closed spacetimes. If there is no null infinity, then all singularities are technically naked.

[edit: Here is a reference showing null infinity for De Sitter space: http://www.math.miami.edu/~galloway/papers/cqg7_11_021.pdf

which suggests my comment about 'closed' needs clarification. ]

This doesn't necessarily have to change GR; you could (I believe) construct the classical limit of such an "evaporation" model by using an SET with a sufficiently large negative pressure. This would violate several energy conditions, so such SETs are usually considered "unphysical", but when quantum effects are included it's no longer clear that the energy conditions always have to hold anyway.

But you would need to add a rule that says any SET the produces a horizon is illegal. I call that modifying GR.

 

[Dale]

everything that people currently have in mind has already been discussed several times and the last discussions appear to not have helped anyone with anything.

I agree. There simply is no substitute for actually learning the math. In the end, discussions on internet forums just can't provide a shortcut.

 

[Dale]

I have to disagree a little bit here. The field equations by themselves describe spacetime dynamics within a region of spacetime. They don't say anything about what regions must exist, do they? So in Schwarzschild coordinates, there is a region of spacetime described by Schwarzschild coordinates

$2GM/c^2 < r < \infty$
$- \infty < t < \infty$
$0 \leq \theta \leq \pi$
$0 \leq \phi < 2 \pi$

The field equations by themselves don't say anything about the existence of other regions. Now, you can argue physically that there should be other regions besides this one, using the principle of geodesic completeness, or by considering how a star collapses, or something. But the field equations themselves don't say what regions of spacetime exist, they only describe how dynamics works within a region. Or at least, it seems that way to me.

Sure, but SC are not the only coordinates, and many of those other coordinates are equally valid solutions of the EFE which do cover regions inside the horizon. Due to the fact that a given chart maps an open subset of the manifold, the existence of any chart covering the interior implies that those events are part of the whole manifold, while the fact that SC doesn't cover them does not imply the opposite.

The only way to get around that is to modify the EFE or impose some sort of ad-hoc restriction to the set of admissible manifolds.

 

[PeterDonis]

The technical definition of horizon uses null infinity.

Yes, I agree. I wasn't disputing your definition of a black hole event horizon; I was only saying that a black hole event horizon can't exist in a spacetime that doesn't have a null infinity. Other kinds of horizons can exist (such as cosmological horizons), but not black hole event horizons.

However, I'm not sure I was right to say that an open (or flat) FRW spacetime doesn't have a future null infinity; I've been trying to find a link to a Penrose diagram of that spacetime but haven't been able to.

my comment about 'closed' needs clarification.

I think the proper term would be "compact", or more precisely "spatially compact"--i.e., any spacelike slice is a compact manifold. Spacelike slices of de Sitter spacetime are, I believe, not compact.

I'm not sure, though, that being spatially compact is equivalent to not having a future null infinity. That's what I think needs further thought.

But you would need to add a rule that says any SET the produces a horizon is illegal. I call that modifying GR.

The rule couldn't be that simple, since a vacuum SET allows a horizon to form. I was actually thinking of something along the lines of: what if it were possible to prove that, when quantum effects are included, the "effective" SET at the classical level is such that a horizon is always prevented from forming (because the closer a horizon comes to forming, the larger the negative pressure is in the effective SET). This wouldn't require modifying the EFE or any of the postulates of GR; it would just be a (rather unexpected, and unlikely in my view, but possible) consequence of how the underlying quantum laws produce an effective SET at the classical level.

 

[PeterDonis]

What I would expect to be the case is that for very large values of $t$, the solutions would settle down to a solution of the unaltered Einstein Field Equations.

That seems reasonable to me.

However, it's not clear to me that you would ever get the interior of a black hole event horizon. So it would settle down to a solution of the EFE that's missing some regions.

Such a solution would be geodesically incomplete, whereas the initial state (Minkowski spacetime) is geodesically complete. So I'm not sure this would work. What I think would happen instead is that you would not be able to construct a solution with a time-varying G that contained the Schwarzschild exterior. See below.

It's sort of like the case with perturbation theory in physics. Certain solutions (bound states for example) can't be obtained perturbatively.

It's true that the maximally extended Schwarzschild solution to the EFE is something like a soliton; I believe some physicists have actually used that term to describe it. That would mean it's not "reachable" as a perturbation of Minkowski spacetime.

I know that seems weird, since it's obviously possible to express the vacuum exterior region as a perturbation of Minkowski spacetime. But that region is not geodesically complete; so the region we're expressing as a perturbation is not a perturbation of *all* of Minkowski spacetime, it's only a perturbation of a *portion* of Minkowski spacetime; in the simplest case, it's the portion of Minkowski spacetime outside some radius r from a chosen central point.

Which leaves the question of what is the initial condition of the region *inside* that radius? If the region inside radius r starts out in a non-vacuum initial state, then the complete initial state is no longer Minkowski spacetime. But if the region inside radius r starts out as vacuum, then as I said above, I don't think you can construct a solution that turns that vacuum interior into a black hole interior by varying G with time; but you could, perhaps, turn that "vacuum" interior (with particles floating around but no gravity) into a non-vacuum interior with a massive gravitating body in it (if the "particles" have enough mass to form such a body once gravity is "turned on").

 

[zonde]

There simply is no substitute for actually learning the math. In the end, discussions on internet forums just can't provide a shortcut.

It depends on what do you mean by "math" and for what purpose do you need it.

 

[TrickyDicky]

I was actually thinking of something along the lines of: what if it were possible to prove that, when quantum effects are included, the "effective" SET at the classical level is such that a horizon is always prevented from forming (because the closer a horizon comes to forming, the larger the negative pressure is in the effective SET).

I have often considered this very "what if", but you have expressed it in a way clearer than anything I've managed to write myself.

This wouldn't require modifying the EFE or any of the postulates of GR; it would just be a (rather unexpected, and unlikely in my view, but possible) consequence of how the underlying quantum laws produce an effective SET at the classical level.

Agreed. But it is refreshing to see that unlikely as it may be it is at least considered as a possibility something (a plausible way of preventing not only singularities but also event horizons from forming) that was not even admitted as a mathematically and/or physically valid scenario in previous discussions. (Even if it was in the literature as shown by PAllen's references by Krauss et al. in the first posts in this thread).

 

[stevendaryl]

what if it were possible to prove that, when quantum effects are included, the "effective" SET at the classical level is such that a horizon is always prevented from forming (because the closer a horizon comes to forming, the larger the negative pressure is in the effective SET). This wouldn't require modifying the EFE or any of the postulates of GR; it would just be a (rather unexpected, and unlikely in my view, but possible) consequence of how the underlying quantum laws produce an effective SET at the classical level.

It's hard for me to see how that could work, because locally, there is nothing that indicates that you're near an event horizon (in the case of a large enough black hole).

 

[PeterDonis]

unlikely as it may be it is at least considered as a possibility something (a plausible way of preventing not only singularities but also event horizons from forming) that was not even admitted as a mathematically and/or physically valid scenario in previous discussions. (Even if it was in the literature as shown by PAllen's references by Krauss et al. in the first posts in this thread).

Just to be clear, it's a "possibility", but I don't think it's given much consideration by mainstream physicists. Just having an SET that violates the energy conditions is not enough, as the rebuttals to the Krauss et al. paper show (the effective SET associated with black hole evaporation violates energy conditions, but the rebuttals show that evaporation by itself can't prevent a horizon from forming). You would need an SET that *grossly* violates the energy conditions, *and* the violation would need to be highly sensitive to how close a horizon was to forming, so to speak--meaning that the violation would need to be highly sensitive to a *nonlocal* property, since locally there is no way to tell how close a horizon is to forming, as stevendaryl pointed out (see my response to him for some further thoughts).

 

[PeterDonis]

locally, there is nothing that indicates that you're near an event horizon (in the case of a large enough black hole).

True, and as I just responded to TrickyDicky, this makes the mechanism I was referring to highly unlikely. But since there are weird nonlocalities in quantum mechanics, I don't think one could make a blanket statement that it is impossible at our current state of knowledge about quantum gravity. Betting odds are another matter, of course.

 

[zonde]

It's hard for me to see how that could work, because locally, there is nothing that indicates that you're near an event horizon (in the case of a large enough black hole).

True, and as I just responded to TrickyDicky, this makes the mechanism I was referring to highly unlikely. But since there are weird nonlocalities in quantum mechanics, I don't think one could make a blanket statement that it is impossible at our current state of knowledge about quantum gravity. Betting odds are another matter, of course.

In GR there is nothing that indicates that you're near an event horizon. And yet globally event horizon have rather observable consequences (black hole).

So if one considers possibility that EH does not form then he has to add some parameter that can indicate nearness of EH. Basically it is gravitational potential that can do that.

And only then one can make speculations like - maybe density of available quantum states goes down as we go down in gravitational potential or anything else like that.

 

[TrickyDicky]

But let's move a bit onto the observational side and away from the math for a little bit.

There's clearly something very massive and rather dark at the center of our galaxy - we can see the orbits of stars around - something.

http://arxiv.org/abs/astro-ph/0210426 "Closest Star Seen Orbiting the Supermassive Black Hole at the Centre of the Milky Way"

I've rescued this post from the beginning of this long thread because I agree its healthy sometimes to move away for a while from the purely mathematical side to what is actually observed, if only to put things in perspective.
It is true that observing stars near the center of our galaxy at Sagittarius A*, orbiting at very high speeds around a common focus is highly suggestive of something very massive there, if we add that this very spot is relatively dark, it is reasonable to suspect there must be "something like a SMBH" there. And it is expected that in a not very long time we'll have more relevant data to help us discern between a black hole or "something else" that no one at this point has a reasonable theory for.

One thing I don't understand very well is that given the huge mass (4.3 million suns) calculated, in a very compact space, why there seems to be no gravitational lensing effects on the stars closest to Sagittarius A*. We do observe this effects in clusters in which the mass is much more disperse.

 

[pervect]

I've rescued this post from the beginning of this long thread because I agree its healthy sometimes to move away for a while from the purely mathematical side to what is actually observed, if only to put things in perspective.
It is true that observing stars near the center of our galaxy at Sagittarius A*, orbiting at very high speeds around a common focus is highly suggestive of something very massive there, if we add that this very spot is relatively dark, it is reasonable to suspect there must be "something like a SMBH" there. And it is expected that in a not very long time we'll have more relevant data to help us discern between a black hole or "something else" that no one at this point has a reasonable theory for.

One thing I don't understand very well is that given the huge mass (4.3 million suns) calculated, in a very compact space, why there seems to be no gravitational lensing effects on the stars closest to Sagittarius A*. We do observe this effects in clusters in which the mass is much more disperse.

I don't know the current status of observations of this, but there is apparently work being proposed.

See for instance http://arxiv.org/abs/1204.2103

The massive black hole at the Galactic center Sgr A* is surrounded by a cluster of stars orbiting around it. Light from these stars is bent by the gravitational field of the black hole, giving rise to several phenomena: astrometric displacement of the primary image, the creation of a secondary image that may shift the centroid of Sgr A*, magnification effects on both images. The near-to-come second generation VLTI instrument GRAVITY will perform observations in the Near Infrared of the Galactic Center at unprecedented resolution, opening the possibility of observing such effects. Here we investigate the observability limits for GRAVITY of gravitational lensing effects on the S-stars in the parameter space [DLS,gamma,K], where DLS is the distance between the lens and the source, gamma is the alignment angle of the source, and K is the source apparent magnitude in the K-band. The easiest effect to be observed in the next years is the astrometric displacement of primary images. In particular the shift of the star S17 from its Keplerian orbit will be detected as soon as GRAVITY becomes operative. For exceptional configurations it will be possible to detect effects related to the spin of the black hole or Post-Newtonian orders in the deflection.

[add]

Another quote may shed some light on the current status

However, the expected astrometric displacements of the primary
image (∼ 20 μas, Gillessen et al. 2009a) and the secondary-image luminosities (K = 20.8 in the
present known best case, Bozza & Mancini 2009) are difficult to detect and the resolution of the
most powerful modern instruments is currently insufficient to perform such high precision astrometry
and photometry.

 

[TrickyDicky]

I don't know the current status of observations of this, but there is apparently work being proposed.

See for instance http://arxiv.org/abs/1204.2103



[add]

Another quote may shed some light on the current status

Ok, thanks. I suspected that the effects were not yet detectable (they might be soon). Actually my comparison with clusters was a little far-fetched, since their masses are orders of magnitude higher.

 

[pervect]

I've come up with a somewhat simpler approach for presenting the solution for the EF geodesic equations.

Let r, v be the Eddington-Finklesein (EF) coordinates, which are presumed to be functions of proper time $\lambda$. Then let:

$$\dot{r} = \frac{dr}{d\lambda} \hspace{3cm} \dot{v} = \frac{dv}{d\lambda}$$

In EF coordinates, $\dot{v} = {dv}/{d\lambda}$ can be interpreted as the redshift from a light source originating at infinity and propagating radially, this is directly observable as the amount of doppler shift of the light source of a known spectral line of the source at infinity.

The Ingoing Eddington Finklestein metric (gemoetrized) is
http://en.wikipedia.org/w/index.php?title=Eddington–Finkelstein_coordinates&oldid=516198830

$$-(1-2m/r) dv^2 + 2\,dv\,dr$$

The Christoffel symbols are:

$$\Gamma^{v}{}_{vv} = \frac{m}{r^2}$$
$$\Gamma^{r}{}_{vv} = \frac{m(1-2m/r)}{r^2}$$
$$\Gamma^{r}{}_{vr} = \Gamma^{r}{}_{rv} = -\frac{m}{r^2}$$So we can write the geodesic equations as

$$\ddot{v} + \Gamma^{v}{}_{vv} \dot{v}^2 = 0$$

$$\ddot{r} + \Gamma^{r}{}_{vv} \dot{v}^2 + 2 \,\Gamma^{r}{}_{vr} \dot{v} \dot{r} = 0$$

It will be convenient to solve for $\dot{r}$ and $\dot{v}$ as a function of r, rather than the proper time $\lambda$To this end, we use the chain rule:

$$\ddot{r} = \frac{d^2 r}{d \lambda^2} = \frac{d \dot{r}}{d r} \frac{d r}{d \lambda} = \frac{d \dot{r}}{d r} \dot{r}$$
$$\ddot{v} = \frac{d^2 v}{d \lambda^2} = \frac{d \dot{v}}{d \lambda} = \frac{d \dot{v}}{dr}\frac{dr}{d\lambda} = \frac{d \dot{v}}{dr} \dot{r}$$

Then we can write the geodesic equations as:

$$\ddot{v} + \Gamma^{v}{}_{vv} \dot{v}^2 = \frac{d \dot{v}}{dr} \dot{r} + \frac{m}{r^2} \dot{v}^2 = 0$$

$$\ddot{r} + \Gamma^{r}{}_{vv} \dot{v}^2 + 2 \,\Gamma^{r}{}_{vr} \dot{v} \dot{r} = \frac{d \dot{r}}{dr} \dot{r} + (1-2m/r)\frac{m}{r^2} \dot{v}^2 - 2 \frac{m}{r^2} \dot{r}\dot{v} = 0$$

Solving this, one way of expressing the general solution is

$$\dot{v} = C\frac{1 \pm \sqrt{2m/r} \sqrt{1-\frac{r-2m}{r_{max} - 2m}}}{1-2m/r}$$
$$\dot{r} = -\frac{m \dot{v}^2}{r^2 \frac{d \dot{v}}{dr}}$$

where C and $r_{max}$ are constants, and the solution has the property that $\dot{r}=0$ at $r=r_{max}$.

Infalling geodesics use the minus sign - this can be confirmed by solving for $\dot{r}$ and taking the limit as r->2m and confirming that only this choice makes $\dot{r}$ negative, which is required for an infalling geodesic.

$$\dot{v} = C\frac{1 - \sqrt{2m/r} \sqrt{1-\frac{r-2m}{r_{max} - 2m}}}{1-2m/r}$$

If we take $r_{max}$ = infinity, and let $\dot{v}$ be unity at infinity, representing a fall from rest at infinity (sometimes called a Lemaitre observer) then we get the expected solution:

$$\dot{v} = \frac{1-\sqrt{2m/r}}{1-2m/r} = \frac{1-\sqrt{2m/r}}{(1-\sqrt{2m/r})(1+\sqrt{2m/r}) } = \frac{1}{1+\sqrt{2m/r}}$$

If we take a finite $r_{max}$ and set $r = r_{max}$, we find that
$$\dot{v}|_{r=r_{max}} = \frac{C}{1-2m/r}$$

so, it appears that the correct value for C is $\sqrt{1-2m/r_{max}}$ as the doppler shift at $r=r_max$ should be the same as the Schwarzschild gravitational time dilation at the same r.

As a partial check on this: the killing vector $\xi^a$ should be $\frac{\partial}{\partial v}$. This makes $\xi_a=(-1+2m/r) \partial v + \partial r$ Since $u^a \xi_a$ should be a constant along the curve, $u^a$ being the four-velocity, (-1+2m/r) \dot{v} + dot{r} should be a constant along the curve. Also, said constant should be equal to the "energy at infinity".

The value of this conserved constant evaluated in the limit as r->2m and again as r->r_max is just C. So we do expect C<1 for a fall from at rest from a finite r_max, and C=1 for a fall from at rest from infinity.

This yields the final expression for the radial doppler shift to infinity as a function of r for a body starting at rest at $r=r_{max}$:

$$\dot{v} =\frac{ \sqrt{1-2m/r_{max}}}{1-2m/r } \left( {1 - \left( \sqrt{2m/r} \right) \left( \sqrt{1-\frac{r-2m}{r_{max} - 2m}} \right) } \right)$$

 

[.Scott]

I have not read all 29 pages of this thread, but I did read the first 3 pages and some of what was posted after that. What I did not find was a full response to the original question as I believe it would unfold.

I'll consider a non-spinning black hole of 100,000 solar masses - easily large enough so that our adventurers will be able to cross the event horizon intact. Of course, this black hole is also completely free of any other falling matter that might interfere with our observations.

I will add "Carol" to the scenario. So we start with Alice, Bob, and Carol all keeping their distance from the event horizon:
1) Alice stops her hover and crosses through the event horizon at exactly 12 noon. Although she can compute the moment she will cross the event horizon, when she passes that point of no possible return, she notices nothing special. However, soon (I'm not sure how soon), unseen by the outside world, tidal forces become extreme and Alice dies from spaghettification.
2) Bob and Carol watch as Alice drops. From their point of view, Alice never quite reached the event horizon and her watch never quite reached 12 noon. As she approaches the event horizon, their ability to measure her momentum with greater and greater precision approaches a potential conflict with the Heisenberg Uncertainly limit and so her position becomes more spread out. She appears more and more holographic.
3) A day later, Bob stops his hover and drops down towards Alice. He soon notices that Carol's hologram stops spreading. Bob drops through the Event Horizon just as his watch reaches noon time of that second day. At that point, he will still see Carol moving down at an unreachable and increasing distance below him. I suspect he will never see her become spaghettified. Instead, he will become distracted by his own spaghettification before the image of Alice's demise ever reaches him.
4) Meanwhile Carol is monitoring the situation. Alice has caused the diameter of the event horizon to increase by a tiny amount. As Bob approaches the event horizon, he obscures Alice's holograph covering it with his own.

 

[ViperSRT3g]

I posted this in a similar topic just a little while ago:

Its speed could be anything you wanted, it's not coming back from beyond the event horizon. (Akin to asking what's outside the universe)

As for objects falling into it, you would see them slowing down, slowly red shifting until it's so red shifted you can't see it anymore.

The reason for this? Imagine the object as it's approaching the black hole's event horizon. It's getting faster and faster approaching c. Throughout this acceleration towards the speed of c, all photons coming from it are already becoming red shifted (they are leaving an object that's moving away from us at nearly the speed of light). Due to relativistic effects, the object is moving so fast, that time is slowing down for it. Thus we see it slowing down, and slowly red shifting.

Now the moment it crosses the event horizon, all photons being emitted from the object will be unable to reach us. So the last thing we see of the object is the object slowing down while red shifting until it's so red shifted you can't see it anymore.

 

[PeterDonis]

What I did not find was a full response to the original question as I believe it would unfold.

Hmm. Your response doesn't include an answer to the question the OP actually asked. See below.

1) Alice stops her hover and crosses through the event horizon at exactly 12 noon. Although she can compute the moment she will cross the event horizon, when she passes that point of no possible return, she notices nothing special.

Ok so far.

However, soon (I'm not sure how soon)

Pretty soon; for a BH of 100,000 solar masses, Alice's proper time to fall from the horizon to the singularity is about 1 second.

unseen by the outside world, tidal forces become extreme and Alice dies from spaghettification.

Ok here, but note that this assumes a classical model of the BH. Quantum effects might change this (and in fact many physicists expect them to). I mention that because you bring in quantum effects later on in your post, but if quantum effects are noticeable at the horizon they should certainly be noticeable near the singularity, yet you haven't taken them into account here. (My understanding of the OP is that it was assuming the classical model, no quantum effects.)

2) Bob and Carol watch as Alice drops. From their point of view, Alice never quite reached the event horizon and her watch never quite reached 12 noon.

Ok here.

As she approaches the event horizon, their ability to measure her momentum with greater and greater precision approaches a potential conflict with the Heisenberg Uncertainly limit and so her position becomes more spread out. She appears more and more holographic.

Only if we add quantum effects to the standard classical model of a BH. (Even then I'm not sure your description matches the mainstream quantum model of infalling objects near the horizon; but that probably deserves a separate thread if you want to discuss it.) In the classical model, light emitted by Alice as she gets closer and closer to the horizon becomes more and more redshifted when it is received by Bob and Carol; but the light just shows Alice getting closer and closer to the horizon without ever quite reaching it.

3) A day later, Bob stops his hover and drops down towards Carol.

You mean towards Alice, correct? (I'll assume so and correct that in what follows.) I'll leave out the "hologram" effects from here on, for the reasons given above.

Bob drops through the Event Horizon just as his watch reaches noon time of that second day. At that point, he will still see [STRIKE]Carol[/STRIKE] Alice moving down at an unreachable and increasing distance below him.

No, he won't; the only light from Alice that he will see at the horizon is the light she emitted when she was at the horizon. To see light she emitted further in, he has to fall further in.

I suspect he will never see her become spaghettified. Instead, he will become distracted by his own spaghettification before the image of [STRIKE]Carol's[/STRIKE] Alice's demise ever reaches him.

This is correct; he won't reach the light that Alice emitted as she was becoming spaghettified until *after* he has become spaghettified himself.

4) Meanwhile Carol is monitoring the situation. Alice has caused the diameter of the event horizon to increase by a tiny amount.

Yes, and Bob causes it to increase by another tiny amount. Carol's observation of Bob is similar to Bob's and Carol's observation of Alice when she fell in; Carol will see more and more redshifted light from Bob, which shows him getting closer and closer to the horizon but never quite reaching it (and always a little bit behind Alice).

However, as I noted above, you left out the question the OP actually asked, which is: when Alice falls in, is there a finite time by Bob's and Carol's clock at which they are no longer able to fly down and rescue Alice before she crosses the horizon? The answer is yes, there is. In fact, we can make an even stronger statement: suppose that Bob has a very strong rope attached to Alice before she starts falling: the rope is so strong that the speed of sound in the rope (which is the speed at which causal influences propagate along it from one end to the other) is equal to the speed of light. Alice starts free-falling towards the hole at time $t = 0$ by Bob's clock. Is there a time by Bob's clock at which he can no longer stop Alice from falling through the horizon by pulling on the rope? Again, the answer is yes, which means that there is some time by Bob's clock at which any light signal he sends inward towards Alice will not reach her until after she has crossed the horizon.

 

[.Scott]

However, soon (I'm not sure how soon),

Pretty soon; for a BH of 100,000 solar masses, Alice's proper time to fall from the horizon to the singularity is about 1 second.

I don't know if it's that quick. Neither Alice nor Bob are really dropping "straight" down as if dropping through a sphere. They're in a TARDIS-like contraption - much bigger on the inside than on the outside. So I think they're free-fall is a kind of speed-of-light slide towards the center.

unseen by the outside world, tidal forces become extreme and Alice dies from spaghettification.

Ok here, but note that this assumes a classical model of the BH. Quantum effects might change this (and in fact many physicists expect them to). I mention that because you bring in quantum effects later on in your post, but if quantum effects are noticeable at the horizon they should certainly be noticeable near the singularity, yet you haven't taken them into account here. (My understanding of the OP is that it was assuming the classical model, no quantum effects.)

Quantum effects in her immediate world would not be seen by Alice as she crossed the event horizon. In fact, if Alice is the only object dropping into the BH, nothing really special should happen to Alice's immediate space until tidal forces start getting uncomfortable.

You mean towards Alice, correct? (I'll assume so and correct that in what follows.)

Thanks for the correction. I've edited my original post.

Bob drops through the Event Horizon just as his watch reaches noon time of that second day. At that point, he will still see Alice moving down at an unreachable and increasing distance below him.

No, he won't; the only light from Alice that he will see at the horizon is the light she emitted when she was at the horizon. To see light she emitted further in, he has to fall further in.

When Bob reaches the event horizon, he will not find Alice there. From Bob's point of view, she will have moved further down - and will be continuing to move further down. I think the main issue here is how much of that he can really "see".

However, as I noted above, you left out the question the OP actually asked, which is: when Alice falls in, is there a finite time by Bob's and Carol's clock at which they are no longer able to fly down and rescue Alice before she crosses the horizon? The answer is yes, there is. ... Again, the answer is yes, which means that there is some time by Bob's clock at which any light signal he sends inward towards Alice will not reach her until after she has crossed the horizon.

That last sentence is the key. If Bob can get a signal to Alice before Alice crosses the EH, then, in theory, something can be done to save her.

 

[PeterDonis]

I don't know if it's that quick.

I do; I calculated it. Most GR textbooks describe the calculation.

Neither Alice nor Bob are really dropping "straight" down as if dropping through a sphere. They're in a TARDIS-like contraption - much bigger on the inside than on the outside. So I think they're free-fall is a kind of speed-of-light slide towards the center.

I'm not sure this is a very good description. I would recommend studying a good GR textbook's treatment of free-fall into a black hole.

Quantum effects in her immediate world would not be seen by Alice as she crossed the event horizon. In fact, if Alice is the only object dropping into the BH, nothing really special should happen to Alice's immediate space until tidal forces start getting uncomfortable.

We don't know for sure that this is true because we don't have a good quantum theory of gravity. Many physicists think it's true, but not all; there are possible quantum models in which it isn't.

When Bob reaches the event horizon, he will not find Alice there.

Yes; I didn't say he would.

From Bob's point of view, she will have moved further down - and will be continuing to move further down.

Yes.

I think the main issue here is how much of that he can really "see".

Which depends on what light signals, emitted outward by Alice, will be received by Bob at what point along Bob's worldline. That's how I determined that Bob won't see Alice being spaghettified before he is spaghettified himself.

If Bob can get a signal to Alice before Alice crosses the EH, then, in theory, something can be done to save her.

As long as it's done within a finite time, by Bob's clock, after Alice starts falling. After that finite time has passed, nothing Bob does can save Alice.

 

[.Scott]

I'm not sure this is a very good description. I would recommend studying a good GR textbook's treatment of free-fall into a black hole.

I read several descriptions. I haven't seen one where the actual computation was done - using the reference frame of the projectile (Alice).
The problem I have is that gravity is going to pull you towards all the mass that entered the black hole before you got there. And all of that mass is a really long distance away. I realize that there is serious time dilation, but I can't see how that dilation will be enough to restore your transit time to seconds.

 

[pervect]

I read several descriptions. I haven't seen one where the actual computation was done - using the reference frame of the projectile (Alice).
The problem I have is that gravity is going to pull you towards all the mass that entered the black hole before you got there. And all of that mass is a really long distance away. I realize that there is serious time dilation, but I can't see how that dilation will be enough to restore your transit time to seconds.

You can find the calculation of proper time to collapse in MTW's "Gravitation", pg 851, for the collapse of an observer on the surface of a sphere of presureless dust that is collapsing into a black hole.

eq 32.10c gives the proper time, in geometric units as $\pi ( R_i^3 / 8M)^ {\frac{1}{2}}$

It might not be done in the way you specify ("using the reference frame of the projectile"), but that doesn't make the computation invalid. One of the points of physics is that you can calculate observations in any convenient coordinate system.

It's a bit unclear to me exactly what you mean when you say "the reference frame of the projectile", or why you think you need to do it in that manner.

 

[PeterDonis]

I read several descriptions. I haven't seen one where the actual computation was done - using the reference frame of the projectile (Alice).

The simplest computation I know of is done in Painleve coordinates, which is about the best you can do for "the reference frame of the projectile" if you need to cover more than a small local patch of spacetime.

And all of that mass is a really long distance away.

Not when you reach the horizon. The time I gave you was the time to fall from the horizon to the singularity. That's because the statement of yours that I was responding to was about how long it takes Alice to fall from the horizon to the singularity (or at least close enough to the singularity to be spaghettified). The time to fall from a long distance away (meaning a long distance above the horizon) will be longer, of course (how much longer depends on how long a distance away--pervect gave the general formula).

 

[.Scott]

I guess it's that singularity that's bothering me.
At any point as you are descending into the BH, you could compute the difference in the circumference of the black hole at your location 1 microsecond earlier and the circumference of the black hole at your current position. Since you are descending towards the mass of the black hole, space curvature should be getting worse and worse so if you weren't accelerating that circumference difference should be getting smaller and smaller. But you are accelerating, and being almost at the speed of light anyway, you would be seeing distances down getting smaller and smaller - allowing the circumference difference to increase.

I'm sure that your equations are right, but when I try to visualize it, I see everything that has ever fallen into the black hole before me accelerating away from me making it harder and harder for me to further reduce the circumference of the black hole at my elevation.

 

[PeterDonis]

At any point as you are descending into the BH, you could compute the difference in the circumference of the black hole at your location 1 microsecond earlier and the circumference of the black hole at your current position.

I don't understand what you mean by this. Are you trying to figure out how much the mass of the BH increases when you yourself fall in? That would be the obvious interpretation of what you are saying here, but you go on to say...

Since you are descending towards the mass of the black hole, space curvature should be getting worse and worse so if you weren't accelerating that circumference difference should be getting smaller and smaller.

...which makes it seem like you mean something else by "circumference of the black hole", something to do with space curvature--which, btw, is different from *spacetime* curvature, so you need to clarify which one you mean.

If you mean space curvature, that is not an invariant; it depends on how you "slice up" the spacetime into space and time. For the most natural way of doing that for an observer free-falling into the hole, the way that corresponds to Painleve coordinates, space curvature is zero--the "space" that the observer finds himself falling through is flat! (Space is curved for the most natural way of slicing spacetime into space and time for a *static* observer--one who "hovers" at a constant altitude above the horizon--and there is a sense in which the "circumference" of this space curvature gets smaller for static observers closer to the horizon. But there are no static observers at or below the horizon.)

If you mean spacetime curvature, that is tidal gravity, and tidal gravity does get stronger as you get closer to the singularity at the center of the hole. But I don't know of any way of describing tidal gravity by a "circumference".

But you are accelerating, and being almost at the speed of light anyway, you would be seeing distances down getting smaller and smaller

No, you wouldn't; at least, not with the most natural slicing of spacetime into space and time for an infalling observer. Space is flat in that slicing, as I noted above, and spatial distances remain the same all the way down. The curved spacetime of a black hole does not work the same way as flat Minkowski spacetime, and you can't carry over all of your intuitions from special relativity.

I'm sure that your equations are right, but when I try to visualize it, I see everything that has ever fallen into the black hole before me accelerating away from me

There is a sense in which this is true: tidal gravity along the radial direction does make objects below you that are also falling inward radially appear to accelerate away from you. However:

making it harder and harder for me to further reduce the circumference of the black hole at my elevation.

Again, I don't understand what this means. See above.

 

[.Scott]

In a normal Euclidean sphere, with no space curvature, the circumference is w=2πr. The ratio 2πr/w = 1.
As space curvature increases, that ratio gets much larger 2πr/w>1.

But if there is no space curvature without acceleration, then Alice will drop until she hits something or begins stretching out.

 

[PeterDonis]

In a normal Euclidean sphere, with no space curvature, the circumference is w=2πr. The ratio 2πr/w = 1.
As space curvature increases, that ratio gets much larger 2πr/w>1.

This kind of space curvature is the kind that depends on how you slice spacetime into space and time. The natural slicing of the infalling observer has no space curvature in this sense; see my previous post.

But if there is no space curvature without acceleration

What does acceleration have to do with it?

then Alice will drop until she hits something or begins stretching out.

A black hole spacetime is vacuum; there's nothing to hit. But I don't see what this has to do with space curvature.

 

[.Scott]

This kind of space curvature is the kind that depends on how you slice spacetime into space and time. The natural slicing of the infalling observer has no space curvature in this sense; see my previous post.

What does acceleration have to do with it?

If not acceleration (for example, resistance to falling), what does create a space curvature?

 

[PeterDonis]

If not acceleration (for example, resistance to falling), what does create a space curvature?

Space curvature is just an aspect of spacetime curvature, which is created by the presence of mass (or more generally, stress-energy). It has nothing to do with "resistance to falling"; I'm not even sure what you mean by that, or how it's equivalent to acceleration, but it doesn't seem like it has anything to do with mass.

 

[.Scott]

Having slept on it, let me try this approach. I'm sure it's a loosing argument, but I don't see the flaw.

First, the event horizon is all that matters. If I doubled the mass of the sun, it would take 9 minutes for the effects to reach Earth, but if I double the mass of a black hole singularity, the effects will never reach the event horizon. From the point of view of someone outside the black hole, the entire mass of the black hole is concentrated on or very near its event horizon.

Giving Alice a 24 hour head start into the 100,000 solar mass black hole means that Bob will see Alice on the surface of a black hole when he starts his drop. But he will never reach Alice. In fact Alice along with the entire mass of the black hole will retreat from him at an ever increasing rate.

With tidal forces being inversely proportional to the square of the distance, spaghettification is fighting an uphill battle - so to speak. My guess is that it would approach a limit - perhaps a survivable limit - leaving Alice and Bob in a never-ending fall. But I wouldn't bet my life on it.

 

[PAllen]

Having slept on it, let me try this approach. I'm sure it's a loosing argument, but I don't see the flaw.

First, the event horizon is all that matters. If I doubled the mass of the sun, it would take 9 minutes for the effects to reach Earth, but if I double the mass of a black hole singularity, the effects will never reach the event horizon. From the point of view of someone outside the black hole, the entire mass of the black hole is concentrated on or very near its event horizon.

Really? When the star was more like a neutron star, matter was spread throughout. Then it collapsed further. You think the matter from the center jumped to near the horizon? Any way in GR of asking what happened to this interior matter concludes it formed a singularity surrounded by an event horizon.

Giving Alice a 24 hour head start into the 100,000 solar mass black hole means that Bob will see Alice on the surface of a black hole when he starts his drop. But he will never reach Alice. In fact Alice along with the entire mass of the black hole will retreat from him at an ever increasing rate.

With tidal forces being inversely proportional to the square of the distance, spaghettification is fighting an uphill battle - so to speak. My guess is that it would approach a limit - perhaps a survivable limit - leaving Alice and Bob in a never-ending fall. But I wouldn't bet my life on it.

Forget 24 hours, but if Alice was hovering just above the horizon, and dropped, while Bob has been falling from far away, timed to cross the horizon just after Alice, the following is definitely possible: Bob will see Alice crossing the horizon when Bob crosses, then Bob will actually catch and pass Alice, reaching the singularity first.

 

[PeterDonis]

If I doubled the mass of the sun

How? You can't just magically double the Sun's mass; that violates the Einstein Field Equation. The added mass has to come from somewhere. The simplest case to analyze mathematically is the case where it falls in as a spherically symmetric shell of matter. For that case, this...

it would take 9 minutes for the effects to reach Earth

...is wrong; Earth will feel the effects as soon as the infalling spherically symmetric shell of matter passes Earth's orbit on the way in. You can construct other scenarios where there will be some time delay between the Sun feeling effects and the Earth feeling effects, but the scenario has to be consistent with the EFE, so mass can't just appear from nowhere.

Much the same point applies to the black hole case:

but if I double the mass of a black hole singularity, the effects will never reach the event horizon.

Yes, they will, because to double the mass of the hole's singularity, the mass has to fall in, which means it has to pass through the horizon first. And someone orbiting the hole will see the effects in much the same way as the Earth would see the effects of a large mass falling into the Sun, as above.

From the point of view of someone outside the black hole, the entire mass of the black hole is concentrated on or very near its event horizon.

No, from the point of view of someone outside the black hole, the mass of the hole is somewhere inside the horizon, but he has no way of telling where. Nor does it matter to him, because it makes no difference to any observations he can make.

Giving Alice a 24 hour head start into the 100,000 solar mass black hole means that Bob will see Alice on the surface of a black hole when he starts his drop.

Huh? The hole has no "surface". If by "surface" you mean "horizon", Bob can't see Alice on the horizon until he himself reaches the horizon; at that instant he will see light that Alice emitted when she crossed the horizon, no matter how long ago.

But he will never reach Alice.

If they are both freely falling in, this is correct; someone who free-falls in later can never catch up to someone who free-falls in earlier. (Things get more complicated if one or both of them can fire rockets to accelerate inward or outward, but I don't think we need to go into that here.)

In fact Alice along with the entire mass of the black hole will retreat from him at an ever increasing rate.

True for Alice, false for the mass of the hole. The mass of the hole never changes at all from Bob's (or Alice's) point of view; and the "location" of the mass--the singularity--isn't a spatial location anyway; it's an instant of time. So it makes no sense for Bob or Alice to ask "how far away" the singularity is; it's in their future. The only question that makes sense is how long it will take them, by their own clock, to reach the singularity. Asking how far away it is would be like asking how far away, in the sense of spatial distance, next Tuesday is.

With tidal forces being inversely proportional to the square of the distance, spaghettification is fighting an uphill battle - so to speak. My guess is that it would approach a limit - perhaps a survivable limit - leaving Alice and Bob in a never-ending fall.

Nope. Tidal forces increase without bound as the singularity is approached, and both Bob and Alice reach the singularity in a finite time by their clocks, meaning that they will be spaghettified in a (slightly shorter) finite time by their clocks.

But I wouldn't bet my life on it.

Good call.

 

[.Scott]

Really? When the star was more like a neutron star, matter was spread throughout. Then it collapsed further. You think the matter from the center jumped to near the horizon? Any way in GR of asking what happened to this interior matter concludes it formed a singularity surrounded by an event horizon.

I was only talking about an existing black hole. I wasn't trying to address exactly what happens when one forms.

Forget 24 hours, but if Alice was hovering just above the horizon, and dropped, while Bob has been falling from far away, timed to cross the horizon just after Alice, the following is definitely possible: Bob will see Alice crossing the horizon when Bob crosses, then Bob will actually catch and pass Alice, reaching the singularity first.

In the scenario created a week or two ago, Alice drops into the black hole first. Bob remains outside the black hole and sees Alice approach the event horizon.
For as long as Bob remains outside the black hole, he can never see Alice cross the event horizon.

Then Bob drops. In that scenario, Bob will never cross paths with Alice. Instead, Alice and everything else that makes up the black hole will fall further and further away. One notion is that Alice will stop when she reaches the singularity, but that singularity doesn't seem reachable. PeterDonis stated there was nothing for Alice to his - just the vacuum of space. If that's true, everyone just keeps falling and the tidal forces should never get extreme.